图像压缩：以像素值存储或传输图像。可以通过减小每个像素包含的值来对其进行压缩。图像压缩基本上有两种类型：
1.无损压缩：在这种类型的压缩中，恢复图像后与使用压缩技术之前的图像完全相同，因此其质量并未降低。
2.有损压缩：在这种类型的压缩中，恢复之后，我们将无法获得与旧数据完全相同的数据，这就是图像质量显着下降的原因。但是，这种类型的压缩会导致图像数据的非常高的压缩率，并且在通过网络传输图像时非常有用。
离散余弦变换用于有损图像压缩中，因为它具有非常强的能量压缩能力，即，其大量信息存储在信号的非常低的频率分量中，而其余的其他频率具有非常小的数据，可以通过使用很少的频率来存储位数(通常最多2或3位)。
要对图像执行DCT变换，首先我们必须获取图像文件信息(像素值，范围为0 – 255的整数)，然后将其分成8 X 8矩阵块，然后对该块进行离散余弦变换。数据。
应用离散余弦变换后，我们将看到其超过90％的数据将包含在低频分量中。为简单起见，我们采用大小为8 X 8的矩阵，所有值均为255(认为图像是完全白色的)，然后将对此进行2-D离散余弦变换以观察输出。
算法：让我们有一个二维变量名为8 x 8的矩阵，它包含图像信息，而一个二维变量dct具有相同的维数，在应用离散余弦变换之后，该二维变量包含信息。所以，我们有公式
dct [i] [j] = ci * cj(sum(k = 0至m-1)sum(l = 0至n-1)矩阵[k] [l] * cos((2 * k + 1)* i * pi / 2 * m)* cos((2 * l + 1)* j * pi / 2 * n)
如果i = 0则ci = 1 / sqrt(m)，否则ci = sqrt(2)/ sqrt(m)和
类似地，如果j = 0则cj = 1 / sqrt(n)，否则cj = sqrt(2)/ sqrt(n)
并且我们必须将此公式应用于所有值，即从i = 0到m-1和j = 0到n-1
在这里， sum(k ＝ 0至m-1)表示从k ＝ 0至k ＝ m-1的值的总和。
在此代码中，m和n都等于8，并且pi定义为3.142857。
C++
// CPP program to perform discrete cosine transform
#include
using namespace std;
#define pi 3.142857
const int m = 8, n = 8;
// Function to find discrete cosine transform and print it
int dctTransform(int matrix[][n])
{
int i, j, k, l;
// dct will store the discrete cosine transform
float dct[m][n];
float ci, cj, dct1, sum;
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
// ci and cj depends on frequency as well as
// number of row and columns of specified matrix
if (i == 0)
ci = 1 / sqrt(m);
else
ci = sqrt(2) / sqrt(m);
if (j == 0)
cj = 1 / sqrt(n);
else
cj = sqrt(2) / sqrt(n);
// sum will temporarily store the sum of
// cosine signals
sum = 0;
for (k = 0; k < m; k++) {
for (l = 0; l < n; l++) {
dct1 = matrix[k][l] *
cos((2 * k + 1) * i * pi / (2 * m)) *
cos((2 * l + 1) * j * pi / (2 * n));
sum = sum + dct1;
}
}
dct[i][j] = ci * cj * sum;
}
}
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
printf("%f\t", dct[i][j]);
}
printf("\n");
}
}
// Driver code
int main()
{
int matrix[m][n] = { { 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 } };
dctTransform(matrix);
return 0;
}
//This code is contributed by SoumikMondal Java
// Java program to perform discrete cosine transform
import java.util.*;
class GFG
{
public static int n = 8,m = 8;
public static double pi = 3.142857;
// Function to find discrete cosine transform and print it
static strictfp void dctTransform(int matrix[][])
{
int i, j, k, l;
// dct will store the discrete cosine transform
double[][] dct = new double[m][n];
double ci, cj, dct1, sum;
for (i = 0; i < m; i++)
{
for (j = 0; j < n; j++)
{
// ci and cj depends on frequency as well as
// number of row and columns of specified matrix
if (i == 0)
ci = 1 / Math.sqrt(m);
else
ci = Math.sqrt(2) / Math.sqrt(m);
if (j == 0)
cj = 1 / Math.sqrt(n);
else
cj = Math.sqrt(2) / Math.sqrt(n);
// sum will temporarily store the sum of
// cosine signals
sum = 0;
for (k = 0; k < m; k++)
{
for (l = 0; l < n; l++)
{
dct1 = matrix[k][l] *
Math.cos((2 * k + 1) * i * pi / (2 * m)) *
Math.cos((2 * l + 1) * j * pi / (2 * n));
sum = sum + dct1;
}
}
dct[i][j] = ci * cj * sum;
}
}
for (i = 0; i < m; i++)
{
for (j = 0; j < n; j++)
System.out.printf("%f\t", dct[i][j]);
System.out.println();
}
}
// driver program
public static void main (String[] args)
{
int matrix[][] = { { 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 } };
dctTransform(matrix);
}
}
// Contributed by Pramod KumarC#
// C# program to perform discrete cosine transform
using System;
class GFG
{
// Function to find discrete cosine transform and print it
public static void dctTransform(int[,] matrix, int m, int n)
{
double pi = 3.142857;
int i, j, k, l;
// dct will store the discrete cosine transform
double[,] dct = new double[m, n];
double ci, cj, dct1, sum;
for (i = 0; i < m; i++)
{
for (j = 0; j < n; j++)
{
// ci and cj depends on frequency as well as
// number of row and columns of specified matrix
if (i == 0)
ci = 1 / Math.Sqrt(m);
else
ci = Math.Sqrt(2) / Math.Sqrt(m);
if (j == 0)
cj = 1 / Math.Sqrt(n);
else
cj = Math.Sqrt(2) /Math.Sqrt(n);
// sum will temporarily store the sum of
// cosine signals
sum = 0;
for (k = 0; k < m; k++)
{
for (l = 0; l < n; l++)
{
dct1 = matrix[k, l] *
Math.Cos((2 * k + 1) * i * pi / (2 * m)) *
Math.Cos((2 * l + 1) * j * pi / (2 * n));
sum = sum + dct1;
}
}
dct[i,j] = ci * cj * sum;
}
}
for (i = 0; i < m; i++)
{
for (j = 0; j < n; j++)
{
Console.Write(dct[i,j]);
}
Console.WriteLine();
}
}
// Driver code
static void Main()
{
const int m = 8, n = 8;
int[,] matrix = new int[m, n] {
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 },
{ 255, 255, 255, 255, 255, 255, 255, 255 }
};
dctTransform(matrix, m, n);
}
}
// This code is contributed by SoumikMondal输出：
2039.999878 -1.168211 1.190998 -1.230618 1.289227 -1.370580 1.480267 -1.626942
-1.167731 0.000664 -0.000694 0.000698 -0.000748 0.000774 -0.000837 0.000920
1.191004 -0.000694 0.000710 -0.000710 0.000751 -0.000801 0.000864 -0.000950
-1.230645 0.000687 -0.000721 0.000744 -0.000771 0.000837 -0.000891 0.000975
1.289146 -0.000751 0.000740 -0.000767 0.000824 -0.000864 0.000946 -0.001026
-1.370624 0.000744 -0.000820 0.000834 -0.000858 0.000898 -0.000998 0.001093
1.480278 -0.000856 0.000870 -0.000895 0.000944 -0.001000 0.001080 -0.001177
-1.626932 0.000933 -0.000940 0.000975 -0.001024 0.001089 -0.001175 0.001298